The program below creates a stream with the numbers 1..995 and then filters the stream, keeping only the even numbers, and then prints the second number in the stream (implemented as the first number of the tail, just like in the 3.5 Section in the Wizard book).
How's this:
Py> from itertools import islice Py> print islice((x for x in xrange(1, 996) if x % 2 == 0), 1, 2).next() 4
Breaking it into pieces:
Py> from itertools import islice Py> stream = (x for x in xrange(1, 996) if x % 2 == 0) Py> second_item = islice(stream, 1, 2).next() Py> print second_item 4
And most of the stream hasn't been consumed yet: Py> print stream.next() 6 Py> unconsumed = list(stream) Py> len(unconsumed) 494
And this version has no problem with recursion limits: Py> print islice((x for x in xrange(1, sys.maxint) if x % 2 == 0), 1, 2).next() 4
(xrange can't handle Python longs, unfortunately, so we *are* constrained by sys.maxint. However, since my machine only has half a gig of RAM, the above is still a damn sight quicker than the equivalent list comprehension would be!)
Something else: this crashes with a "maximum recursion reached" . print stream_enumerate_interval(1,998)
while this does not crash . print stream_enumerate_interval(1,900) this means Python has a maximum of something like 900 recursions?
The CPython implementation is limited by the stack size allocated by the C runtime library. The exact recursion limit is platform dependent, but something around 1000 sounds fairly normal.
Cheers, Nick.
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Nick Coghlan | [EMAIL PROTECTED] | Brisbane, Australia
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http://boredomandlaziness.skystorm.net
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